Problem: Simplify the following expression: $y = \dfrac{7x^2+32x- 15}{7x - 3}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(7)}{(-15)} &=& -105 \\ {a} + {b} &=& &=& {32} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-105$ and add them together. Remember, since $-105$ is negative, one of the factors must be negative. The factors that add up to ${32}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-3}$ and ${b}$ is ${35}$ $ \begin{eqnarray} {ab} &=& ({-3})({35}) &=& -105 \\ {a} + {b} &=& {-3} + {35} &=& 32 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({7}x^2 {-3}x) + ({35}x {-15}) $ Factor out the common factors: $ x(7x - 3) + 5(7x - 3)$ Now factor out $(7x - 3)$ $ (7x - 3)(x + 5)$ The original expression can therefore be written: $ \dfrac{(7x - 3)(x + 5)}{7x - 3}$ We are dividing by $7x - 3$ , so $7x - 3 \neq 0$ Therefore, $x \neq \frac{3}{7}$ This leaves us with $x + 5; x \neq \frac{3}{7}$.